What is an efficient way to measure similarity between two strings? (Levenshtein Distance makes stack too deep)

问题

So, I started with this: http://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance#Ruby

Which works great for really small strings. But, my strings can be upwards of 10,000 characters long -- and since the Levenshtein Distance is recursive, this causes a stack too deep error in my Ruby on Rails app.

So, is there another, maybe less stack intensive method of finding the similarity between two large strings?

Alternatively, I'd need a way to make the stack have much larger size. (I don't think this is the right way to solve the problem, though)

回答1:

Consider a non-recursive version to avoid the excessive call stack overhead. Seth Schroeder has an iterative implementation in Ruby which uses multi-dimensional arrays instead; it appears to be related to the dynamic programming approach for Levenshtein distance (as outlined in the pseudocode for the Wikipedia article). Seth's ruby code is reproduced below:

def levenshtein(s1, s2)
  d = {}
  (0..s1.size).each do |row|
    d[[row, 0]] = row
  end
  (0..s2.size).each do |col|
    d[[0, col]] = col
    end
  (1..s1.size).each do |i|
    (1..s2.size).each do |j|
      cost = 0
      if (s1[i-1] != s2[j-1])
        cost = 1
      end
      d[[i, j]] = [d[[i - 1, j]] + 1,
                   d[[i, j - 1]] + 1,
                   d[[i - 1, j - 1]] + cost
                  ].min
      next unless @@damerau
      if (i > 1 and j > 1 and s1[i-1] == s2[j-2] and s1[i-2] == s2[j-1])
        d[[i, j]] = [d[[i,j]],
                     d[[i-2, j-2]] + cost
                    ].min
      end
    end
  end
  return d[[s1.size, s2.size]]
end

来源：https://stackoverflow.com/questions/8619785/what-is-an-efficient-way-to-measure-similarity-between-two-strings-levenshtein